Python numpy bug or feature

Question

Is this a bug or a feature?

import numpy as np a=b=c=0 print 'a=',a print 'b=',b print 'c=',c a = 5 print 'a=',a print 'b=',b print 'c=',c b = 3 print 'a=',a print 'b=',b print 'c=',c x=y=z=np.zeros(5) print 'x=',x print 'y=',y print 'z=',z x[2]= 10 print 'x=',x print 'y=',y print 'z=',z y[3]= 20 print 'x=',x print 'y=',y print 'z=',z

The output of the code shows me that the numpy initializations are clones of each other while python tends to treat them as independent variable.

a= 0 b= 0 c= 0 a= 5 b= 0 c= 0 a= 5 b= 3 c= 0 x= [ 0. 0. 0. 0. 0.] y= [ 0. 0. 0. 0. 0.] z= [ 0. 0. 0. 0. 0.] x= [ 0. 0. 10. 0. 0.] y= [ 0. 0. 10. 0. 0.] z= [ 0. 0. 10. 0. 0.] x= [ 0. 0. 10. 20. 0.] y= [ 0. 0. 10. 20. 0.] z= [ 0. 0. 10. 20. 0.]

I hope the problem is clear. Is this a bug or a feature in numpy?

Regards

avasal · Accepted Answer · 2012-07-16 09:21:54Z

this is not a bug,, and it is not about numpy initialization, this is a python thing,, check id of both x,y & z in your case, they point to same element

What your code is doing is multiple initialization in the same line, when this happens, only 1 object is created and all the variables refer to the same.

See the below example, how rebinding helps...

In [19]: a=b=[1,2,3] In [20]: a Out[20]: [1, 2, 3] In [21]: b Out[21]: [1, 2, 3] In [22]: a[1] Out[22]: 2 In [23]: a[1] = 99 In [24]: a Out[24]: [1, 99, 3] In [25]: b Out[25]: [1, 99, 3] In [26]: id(a) Out[26]: 27945880 In [27]: id(b) Out[27]: 27945880 In [28]: a = a[:] # This is Rebinding In [29]: a Out[29]: [1, 99, 3] In [30]: id(a) Out[30]: 27895568 # The id of the variable is changed

lvc · Accepted Answer · 2012-07-16 09:17:13Z

This isn't a numpy thing, its a standard Python thing. The same will happen with lists:

>>> a = b = [] >>> a.append(5) >>> a [5] >>> b [5] >>> a[0] = 10 >>> a [10] >>> b [10]

When you do this:

>>> a = 5

You are rebinding the name 'a' to a different object - but when you do slice assignment, you're modifying part of the existing object in place.

Tim Pietzcker · Accepted Answer · 2012-07-16 09:16:57Z

This is not numpy's problem, this is a typical Python feature: Everything is an object, but some objects are mutable and some aren't.

So if you do x=y=z=["foo", "bar"], you bind the exact same object to the three variables. This means that if you change x by mutating the list it references, you change the object that y and z are pointing to, too.

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